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n200080 [17]
2 years ago
6

The mass percent of oxygen in pure glucose, c6h12o6 is 53.3 percent. a chemist analyzes a sample of glucose that contains impuri

ties and determines that the mass percent of oxygen is 49.7 percent. which of the follow impurities could account for the low mass percent of oxygen in the sample?
Chemistry
1 answer:
Dafna1 [17]2 years ago
7 0
We can calculate the mass percent of an element by dividing its atomic mass by the mass of the compound and then multiply by 100:
     % by mass of element = (mass of element/mass of compound) x100%
Impurities like n-eicosane with the molecular formula C20H42 could account for the low percent by mass of oxygen in the sample because it has a zero percent oxygen based on its compound formula which indicates that it does not have the element oxygen.
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Iodine has a lower atomic weight than tellurium (126.90 for I, 127.60 for Te) even though it has a higher atomic number (53 for
andrew11 [14]

Answer and Explanation:

Iodine have lower atomic mass than tellurium even though the atomic number of iodine is more than the atomic number of tellurium

This is because the atomic weight of any element is the sum of number of proton and number of neutron, even though the number of proton in iodine is more so but the number of neutron is less as compared to tellurium which makes the tellurium of high atomic mass

6 0
2 years ago
Read 2 more answers
Choose the correct description for each of the
Aliun [14]

Answer:

Density: Physical Property

Flammability: Chemical Property

Solubility In Water: Physical Property

Reactivity With Water: Chemical Property

Melting Pot: Physical Property

Color: Physical Property

Odor: Physical Property

Explanation:

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7 0
2 years ago
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Which solution contains the largest number of moles of chloride ions?
elixir [45]
Molarity = number of moles of solute/liters of solution
number of moles of solute = molarity x liters of solution

Part (a): <span>30.00 ml of 0.100m Cacl2
number of moles of CaCl2 =  0.1 x 0.03 = 3x10^-3 moles
1 mole of CaCl2 contains 2 moles of chlorine, therefore 3x10^-3 moles of CaCl2 contains 6x10^-3 moles of chlorine

Part (b): </span><span>10.0 ml of 0.500m bacl2
number of moles of BaCl2 = 0.5 x 0.01 = 5x10^-3 moles
1 mole of BaCl2 contains 2 moles of chlorine, therefore 5x10^-3 moles of BaCl2 contains 10x10^-3 moles of chlorine

Part (c): </span><span>4.00 ml of 1.000m nacl
number of moles of NaCl = 1 x 0.004 = 0.004 moles
1 mole of NaCl contains 1 mole of chlorine, therefore 4x10^-3 moles of NaCl contains 4x10^-3 moles of chlorine

Part (d): </span><span>7.50 ml of 0.500m fecl3
number of moles of FeCl3 = 0.5 x 0.0075 = 3.75x10^-3 moles
1 mole of FeCl3 contains 3 moles of chlorine, therefore 3.75x10^-3 moles of FeCl3 contains 0.01125 moles of chlorine

Based on the above calculations, the correct answer is (d)</span>
5 0
2 years ago
Read 2 more answers
Calculate the number of chloride ions in 6.8 g of zinc chloride
Bumek [7]

Answer:

The correct answer is 0.300 * 10^23 ions.

Explanation:

Based on the given question, there is a need to find the number of chloride ions in the mentioned 6.8 grams of zinc chloride compound.  

The moles of zinc chloride (ZnCl2) is,  

= mass of zinc + 2 mass of chlorine

= 65.38 + 2 (35.45)

=65.38 + 70.90

= 136.28 grams (The molecular mass of zinc is 65.38 and the molecular mass of chlorine is 35.45)

Thus, 136.28 g of ZnCl2 contains 70.90 grams of chlorine

Therefore, 6.8 grams of ZnCl2 will comprise = (70.90/136.28) * 6.8

= 3.537 g of chlorine

70.90 g of Cl comprise 6.022*10^23 chlorine, thus, 3.537 g of Cl will comprise (6.022*10^23/70.90) * 3.537

= 0.300 * 10^23 ions of chlorine.  

3 0
2 years ago
If 36.0 g of NaOH (MM = 40.00 g/mol) are added to a 500.0 mL volumetric flask, and water is added to fill the flask, what is the
Kobotan [32]

Answer:

Molarity of NaOH = 1.8 M.

Explanation:

From the question given above, the following data were obtained:

Mass of NaOH = 36 g

Molar mass of NaOH = 40 g/mol

Volume = 500 mL

Molarity of NaOH =?

Next, we shall determine the number of mole in 36 g of NaOH. This can be obtained as follow:

Mass of NaOH = 36 g

Molar mass of NaOH = 40 g/mol

Mole of NaOH =?

Mole = mass / molar mass

Mole of NaOH = 36 / 40

Mole of NaOH = 0.9 mole

Next, we shall convert 500 mL to L. This can be obtained as follow:

1000 mL = 1 L

Therefore,

500 mL = 500 mL × 1 L / 1000 mL

500 mL = 0.5 L

Finally, we shall determine the molarity of NaOH. This can be obtained as follow:

Mole of NaOH = 0.9 mole

Volume = 0.5 L

Molarity of NaOH =?

Molarity = mole / Volume

Molarity of NaOH = 0.9 / 0.5

Molarity of NaOH = 1.8 M

8 0
1 year ago
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